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Question Number 65386 by mathmax by abdo last updated on 29/Jul/19

$$calculate\sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k-1}}{99k-1}$$

Commented by mathmax by abdo last updated on 31/Jul/19

$$letI={\int }_1^{+\mathrm{\infty }}\frac{dx}{1+x^{99}}cha7gementx=\frac{1}{t}give$$$$I=-{\int }_0^1\frac{1}{1+\frac{1}{t^{99}}}(-\frac{dt}{t^2})={\int }_0^1\frac{t^{99}}{t^2(1+t^{99})}dt={\int }_0^1\frac{t^{97}}{1+t^{99}}dt$$$$={\int }_0^1{t}^{97}\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{t}^{99k}dt=\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{\int }_0^1{t}^{99k+97}dt$$$$=\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{\left[\frac{1}{99k+98}{t}^{99k+98}\right]}_0^1=\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^k}{99k+98}$$$${=}_{k=p-1}\sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^{p-1}}{99(p-1)+98}=\sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^{p-1}}{99p-1}\Rightarrow$$$$\sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k-1}}{99k-1}={\int }_1^{+\mathrm{\infty }}\frac{dx}{1+x^{99}}$$$${\int }_1^{+\mathrm{\infty }}\frac{dx}{1+x^{99}}={\int }_1^0(....)dx+{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^{99}}={\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^{99}}-{\int }_0^1\frac{dx}{1+x^{99}}$$$$changementx=t^{\frac{1}{99}}give{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^{99}}={\int }_0^{\mathrm{\infty }}\frac{1}{1+t}\frac{1}{99}{t}^{\frac{1}{99}-1}dt$$$$=\frac{1}{99}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{99}-1}}{1+t}dt=\frac{1}{99}\frac{\pi }{sin\left(\frac{\pi }{99}\right)}=\frac{\pi }{99sin\left(\frac{\pi }{99}\right)}$$$${\int }_0^1\frac{dx}{1+x^{99}}={\int }_0^1\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^{99n}dx=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^1{x}^{99n}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\left[\frac{1}{99n+1}{x}^{99n+1}\right]}_0^1=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{99n+1}=S$$$$=1-\frac{1}{100}+\frac{1}{199}-\frac{1}{3.99+1}+....\Rightarrow$$$$\sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k-1}}{99k-1}=\frac{\pi }{99sin\left(\frac{\pi }{99}\right)}-S$$

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